To try 5 management wants an estimate of the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: rom all the newspapers printed during a single day. In this sample, 35 contain some type of nonconforming attribute. Construct and interpret a 90% confidence interval for the proportion of newspapers printed during the day that have a nonconforming attribute. To Try… 5. Management wants an estimate of the proportion of the corporation’s employees who favour a bonus plan. From a random sample of 344 employees, it was found that 261 were in favour of this particular plan. Find a 90% confidence interval estimate of the population proportion that favours this modified bonus plan. p α=.10; zα/2 = z0.05 =1.645; Confidence interval: p ± zα 2 p( − p) 1 n = 261/344 = 0.7587; n = 344 Sample: 344 proportion p­bar: 0.75872093 significance α: 0.1 confidence: 0.7587⋅ 0.2413 344 0.7587 ± 0.0379 size n: 90% standard error of the proportion: 0.023068621 0.7587 ± 1.645 Interval Estimation Sampling Distribution: 0.7208 ≤ p ≤ 0.7966 Or using MS Excel: Interval Estimate: Lower: =.7587– NORM.S.INV(1–.10/2)*SQRT(.7587*(1–.7587)/344)=0.720754373 zα/2: 1.644853627 margin of error: 0.037944505 To Try… The operations manager at a large newspaper wants to estimate the proportion of newspapers printed that have a nonconforming attribute (e.g., excessive ruboff, improper page setup, missing or duplicate pages). A random sample of 200 newspapers is selected from all the newspapers printed during a single day. In this sample, 35 contain some type of nonconforming attribute. Construct and interpret a 90% confidence interval for the proportion of newspapers printed during the day that have a nonconforming attribute. p α=.10; zα/2 = z0.05 =1.645; = 35/200 = 0.175; n = 200 6. Confidence interval: p ± zα 2 p (1− p) n Sample: size n: 200 proportion p­bar: 0.175 significance α: 0.1 confidence: 90% Lower: =.175– Interval Estimate: NORM.S.INV(1–.10/2)*SQRT(.175*(1–.175)/200)=0.130806514 standard error of the proportion: 0.026867732 0.175 ± 1.645 0.175⋅0.825 200 Interval Estimation 0.175 ± 0.0442 0.1308 ≤ p ≤ 0.2192 Sampling Distribution: Or using MS Excel: Upper: zα/2: 1.644853627...
View Full Document

Ask a homework question - tutors are online